Chapter 5. Entropy, single-orbit algebra, and cancellation

This chapter follows Proposition 6.1, Lemmas 6.3–6.5, and Section 7 of R (pp. 33–38). The inputs are the original finite-alphabet zero-entropy mixing process and the measurable finite-block sums from Chapter 4. The proof first obtains cancellation along each abstract label orbit in an arbitrary fixed coupling. Bounded convergence then contradicts the constant positive integral. We distinguish the zero-entropy estimate, the algebraic alternatives, and the differencing argument that needs only mixing.

Fixed source, from page 33

1. The input in one fixed fibre

Set aside the parameter $u$ and fix a probability space $(Y,\eta )$. Assume only a measurable map $x:Y\to X$ with $x_*\eta =\mu _X$. No time action on $Y$ is required; in particular, there is no assumption of invariance of $\eta$, mixing of $Y$, or zero entropy of $Y$.

The system $X$ is a two-sided stationary finite-alphabet process with time map $T_X$. Its alphabet partition has zero entropy rate, and $T_X$ is mixing. Fix a centered, finite-valued one-site function $f$: the function $f$ depends only on the zeroth symbol of $X$.

Also given are a countable abelian group $L$, an automorphism $G$, and measurable functions $\chi _l:Y\to \mathbb C$ satisfying equation (60) of the source:

The constant phase in a product may depend on $l,l'$ but not on $y$. Since $\chi _l\chi _{-l}$ is a constant phase, $\overline{\chi _l}$ is a constant-phase multiple of $\chi _{-l}$. Representatives of distinct labels are therefore orthogonal.

Proposition 6.1 asserts that, for every fixed $l\in L$,

The absolute value surrounds each individual integral, inside the average. Cancellation only in the signed or complex average would not suffice for Proposition 5.5.

Two mechanisms give this cancellation. Dissociated labels are controlled by the zero entropy of alphabet words. Labels that become group-valued polynomials on arithmetic progressions are controlled by ordinary mixing and Hilbert-space differencing. The second mechanism does not use zero entropy.

2. Lemma 6.3: deterministic coefficients first

On pp. 34–35, a label sequence $(l_m)$ is called dissociated if, for every nonempty finite set of distinct indices $A\subseteq\{0,1,2,\ldots\}$ and every choice $\varepsilon_m\in\{-1,1\}$, one has $\sum_{m\in A}\varepsilon_m l_m\ne0$. Repeated indices are excluded from the definition. It rules out zero labels and repeated labels, but allows torsion in the group.

Take arbitrary deterministic complex coefficients $w_m$ with $|w_m|\le 1$, and choose $0<\theta \le 1$. For a real number $t$, the Taylor remainder satisfies $|\mathrm{e}^{t}-1-t|\le (\mathrm{e}/2)t^{2}\le 2t^{2}$ when $|t|\le 1$. Thus $\mathrm{e}^{t}\le 1+t+2t^{2}$, and

The right-hand side is nonnegative, so the inequalities can be multiplied over $m<n$. On expanding and integrating, each nonconstant term takes one factor $\chi _{l_m}$ or its conjugate from each of a nonempty set of distinct indices. The product rule, allowing constant phases, makes it a constant-phase multiple of $\chi_{\sum_{m\in A}\varepsilon_m l_m}$. Dissociation makes this label nonzero, so its integral vanishes. Only the term choosing constants everywhere survives. Here $A$ is the set of indices selected in the nonconstant term. Consequently,

No independence of the $\chi _{l_m}$ is needed. Their finite-product integral rule suffices. Torsion causes no omission, because each term uses coefficients $\pm 1$ at distinct indices only.

Fix $0<\varepsilon \le 1$ and take $\theta =\varepsilon /4$. Markov's inequality gives

The estimate is uniform over all deterministic coefficients $w_m$. It cannot be applied directly to arbitrary coefficients depending on $y$. Handling such $w_m$ is the next step.

3. Random words without conditioning the label law

Now let $F_m$ take values in a fixed finite subset of the unit disk, with arbitrary coupling to all the labels. Assume only that the Shannon entropy of the random word $W_n=(F_0,\ldots ,F_{n-1})$ satisfies $H_n=o(n)$. The sequence $(F_m)$ need not be stationary or independent.

Put

Call a word likely if its probability is at least $\exp(-a_n n)$. There are at most $\exp(a_n n)$ such words, since their probabilities sum to at most one. The self-information $-\log P(W_n)$ has expectation $H_n$, so the total probability of unlikely words is at most

When $H_n=0$, the ratio is zero. Also, $a_n\to 0$.

Fix arbitrary deterministic phases $\omega_m\in\mathbb T$ and define

On the event that $W_n$ equals a fixed likely word $v=(v_m)$, the variable $Y_n$ is the deterministic-coefficient expression in (63) with $w_m=\omega_m v_m$. Hence

This is an inclusion of events. The probability on the right is not replaced by a probability conditional on $W_n=v$, which could alter the label law, its means, and all the product integrals. A union bound over likely words gives

First fix $\varepsilon$, then let $n\to \infty$. Since $a_n\to 0$, eventually $a_n\le \varepsilon ^{2}/16$, and the exponential term is at most $\exp(-\varepsilon ^{2}n/16)$. This is the estimate that allows an arbitrary coupling.

Finally, for each $m$ choose a deterministic phase $\omega _m$ so that

is real and nonnegative; if the integral vanishes, choose any phase. These phases are numbers determined by the integrals, not functions of $y$. Thus $|Y_n|\le 1$, and

Let $n\to \infty$ and then $\varepsilon \downarrow 0$ to prove Lemma 6.3.

Where zero entropy of the original process is used

For fixed $p\ge 1$ and $0\le j<p$, take $F_m=f(T_X^{pm+j}x)$, first scaling $f$ so that $\|f\|_{\infty }\le 1$. This random word is a function of a consecutive alphabet window in the original $X$, of length at most $pn+p$. If the original window entropy is $H_X(L)=o(L)$, then

because $p,j$ are fixed and $x_*\eta =\mu _X$. Stationarity shifts the window to the coordinate interval used in the definition of entropy rate.

No entropy of $Y$ is computed, and no time action is imposed on $\eta$. Arbitrary couplings of the fibres with the labels leave the alphabet-word law in their $X$ marginal unchanged.

4. Lemma 6.4: algebraic alternatives for one orbit

On pp. 35–36, the alternatives are: the full orbit is dissociated; or each suborbit in finitely many fixed residue classes is dissociated; or each such suborbit is a group-valued polynomial. The alternatives need not be mutually exclusive.

4.1 A finite $\pm 1$ relation gives finite generation in both directions

If $(G^{r}l)_{r\ge 0}$ is not dissociated, take a nonzero finite $\pm 1$ relation. Remove its initial zero coefficients and apply an inverse power of $G$ to write it as

If $q=0$, then $l=0$, which is already the polynomial case. Suppose $q\ge 1$. The leading coefficient is a unit $\pm 1$, so $G^ql$ is an integer linear combination of $l,\ldots ,G^{q-1}l$. Shifting the relation recursively generates all subsequent forward iterates. The trailing coefficient is also $\pm 1$, allowing the backward iterates to be expressed recursively. Thus the two-sided orbit group

is generated by the $q$ elements $l,\ldots ,G^{q-1}l$. Without unit integer coefficients at both ends, this argument would not give finite generation in both directions.

By the structure theorem for finitely generated abelian groups, the group $K$ has finite torsion subgroup $K_{\mathrm{tor}}$, and the quotient $\overline K$ is a free lattice of finite rank. The map $G$ induces its integer lattice automorphism $\overline G$.

4.2 An expanding eigenvalue gives dissociation in every residue class

If $\overline G$ has an eigenvalue satisfying $|\lambda |>1$ on the complexified vector space, choose a nonzero left eigenfunctional $\varphi$ with $\varphi\overline G=\lambda\varphi$. The two-sided orbit of $\overline l$ spans that space, so $\varphi (\overline l)\ne 0$. Otherwise $\varphi$ would vanish on the entire orbit and would be zero.

Choose $p$ so that $R=|\lambda |^{p}>2$. For every $j\in\{0,\ldots,p-1\}$,

A relation at distinct $m$ indices with finitely many coefficients $\pm 1$ would have a largest index $M$. Its largest term has modulus greater than the sum of all preceding terms, because

This is a contradiction. Every residue-class suborbit is therefore dissociated, with the same $p$ for all finitely many $j$. Excluding a relation in a linear image of the quotient also excludes it in $K$; the torsion subgroup need not vanish.

4.3 Without expansion, nilpotent differences must still account for torsion

If $\overline K$ has positive rank and no eigenvalue has modulus greater than one, the identity $\det\overline G=\pm1$ forces every eigenvalue to have modulus one. Their being roots of unity follows from the following direct argument.

The characteristic polynomial of $\overline G^{n}$ has integer coefficients. In fixed dimension $d$, its coefficient of order $a$ is formed from $d$ numbers of modulus one by taking, up to sign, the elementary symmetric expression of degree $a$. Its absolute value is at most $\binom da$, independently of $n$. Only finitely many such characteristic polynomials are possible. For each $\lambda$, all positive powers $\lambda ^{n}$ are roots of these finitely many polynomials, so they take only finitely many values. Two powers coincide, making $\lambda$ a root of unity.

Choose $p$ so that all eigenvalues of $\overline G^{p}$ equal one and $G^{p}$ acts identically on the finite group $K_{\mathrm{tor}}$. Put $N=G^{p}-\operatorname{id}$. There exists $d_{0}\ge 1$ such that

The first inclusion follows from finite-dimensional nilpotence and the embedding of the torsion-free quotient lattice into its complexification. The second follows from the choice of $p$. Hence

The extra application of $N$ cannot simply be omitted, since no free complement invariant under $G$ has been chosen. For example, take $K=\mathbb Z \oplus \mathbb Z /2$ and $G(n,e)=(n,e+n\bmod2)$. The action is the identity on both the free quotient and the torsion subgroup, but $G-\operatorname{id}$ itself need not vanish; its square does.

If $\overline K$ has rank zero, then $K$ is finite. Taking the order of $G$ on $K$ as $p$ gives $N=0$, covering this case as well. Choose a positive integer $d$ with $N^d=0$. The integer binomial expansion gives

The labels in each residue class are thus a group-valued polynomial. Integer binomial coefficients make the identity meaningful even in a group with torsion; no division is used.

5. Lemma 6.5: polynomial labels require only mixing

On pp. 36–37, $X$ need no longer have zero entropy or a finite alphabet. Assume only that $X$ is mixing, that $x_*\eta =\mu _X$, and that the label relations (60) hold. Take any fixed $p\ge 1$, any bounded centered function $F$, and a group-valued polynomial

The assertion is

The polynomial $P(m+h)-P(m)$ has degree at most $d-1$. Indeed, Vandermonde's identity expresses $\binom{m+h}a-\binom ma$ as an integer linear combination of $\binom mb$ with $b<a$. This degree reduction works in any abelian group.

5.1 The constant-label base case

If $P(m)=l_0$, choose $g\in L^\infty(\mu_X)$ satisfying $g\circ x=\mathbb E[\overline{\chi_{l_0}}\mid\sigma(x)]$, with $|g|\le1$. Then

by mixing and $\int F=0$. First obtain convergence of individual terms, then take the Cesàro average of their absolute values. This proves the base case, including $l_{0}=0$.

The function $g$ may depend on the coupling. No mixing rate uniform over all possible $g$ is needed. Eventually this result will be applied separately in each fibre.

5.2 Phase alignment and differencing in one fixed $L^{2}(\eta )$

Assume the conclusion for every lower degree. Choose deterministic phases $\omega _m$ aligning the current integrals, and define, in the fixed Hilbert space $L^{2}(\eta )$,

We have $\|v_m\|_{2}\le \|F\|_{\infty }$. For a fixed positive integer $h$, the second-variable-linear inner product and the label product relation up to constant phase give

Both the alignment phases and the constant phases from multiplication disappear inside the absolute value. Split $F_h$ into its centered part $F_h-\int F_h$ and its constant part. The induction hypothesis applies to the first at lower degree; each integral from the second has absolute value at most $|\int F_h|$. Thus

Fix $h$ before taking $n\to \infty$. As $n$ varies, $h$ is not allowed to vary inside that inductive limit.

5.3 Completing the Hilbert-space van der Corput estimate

Fix $H\ge 1$. Write $A_n=n^{-1}\sum_{m=0}^{n-1}v_m$ and replace it by

At most $H$ boundary terms are removed and added, giving $\|A_n-B_n\|\le 2H\|F\|_{\infty }/n$.

Convexity of the squared norm gives

Expand the square. The $H$ diagonal terms contribute at most $\|F\|_{\infty }^{2}/H$. For positive lag $h$ there are $H-h$ pairs, and the reverse pairs have the same real part. For every fixed $a,h$, shifting the initial index of the average changes at most $2a$ inner products, each of modulus at most $\|F\|_{\infty }^{2}$, so the error tends to zero. Using (66) gives

Mixing and centering imply

After taking $n\to \infty$, let $H\to \infty$. Both the diagonal term and the weighted Cesàro term tend to zero. Thus $A_n$ tends to zero in $L^{2}(\eta )$.

The phases have been aligned, so

This completes the induction. Time acts only on the $X$ variable; there is no change of variables in $\eta$ under a time map. Averaging $v_{m+a}$ shifts sequence indices in a fixed Hilbert space. It does not assert the existence of a transformation preserving $\eta$.

6. Combining residue classes to prove Proposition 6.1

Apply Lemma 6.4 to the fixed label $l$.

• If the full orbit is dissociated, apply the zero-entropy lemma directly with $F_m=f(T_X^mx)$.
• If a fixed $p$ makes every suborbit $G^{pm+j}l$ dissociated, apply the alphabet-window entropy estimate from Section 3 and the zero-entropy lemma for each $j$.
• If each suborbit is the group-valued polynomial in (65), apply Lemma 6.5 with $F=f\circ T_X^{j}$. Invariance of $\mu _X$ keeps it centered. Lemma 6.5 allows an arbitrary bounded $F$, which need not be a function of the zeroth symbol alone.

Write each absolute correlation as $a_r\ge 0$ with $a_r\le \|f\|_{\infty }$. For fixed $p$, split the first $N$ terms according to $r=pm+j$. Residue class $j$ contains $n_j(N)$ terms, with $n_j/N\to 1/p$ and $n_j\to \infty$. Thus, for $N\ge p$,

Each of the finitely many residue-class averages tends to zero, proving (61). The parameters $p,d$ may depend on the label $l$. No rate uniform over every group label is needed.

7. The integral contradiction in Section 7 of R

Now return to the parameter fibres of Proposition 5.5. There is a fixed finite set $B\subseteq L$, one common abstract group automorphism $G$, a fixed centered one-site function $f$, and the measurable functions defined below. We now denote the abstract group $\mathcal L$ from Chapter 4 by $L$ and abbreviate $\kappa (y)$ as $x(y)$, following the notation in Section 6 of the source.

On a common conull set they satisfy

Measurability of these $s_r$ was established through the intrinsic moving-block coefficient sum in (58) of Chapter 4. Only the set identity $J_u(G^{r}B)=B_{S_{0}^{-r}b}(u)$ is used here. The labelwise equivariance identity $\tau _uJ_u=J_{T_Uu}G^{-1}$ for every $u$ is a stronger, unproved assertion that would make the individual-label time rule constant under the identifications; it is unnecessary for this argument.

The equality of versions and the bound hold for each $r$. Intersecting over the countably many $r\ge 0$ gives the common conull set above. The individual-label formula involving the possibly nonmeasurable $J_u$ cannot replace the earlier measurability proof for $s_r$.

Fix a typical $u$. The marginal of $x$ under $\eta _u$ is exactly $\mu _X$, and the label functions satisfy (60) in this fibre. Apply Proposition 6.1 separately to the finitely many $l\in B$ to obtain

This is a fibrewise application, with no invariance, zero-entropy, or mixing assumption on $\eta _u$. The phase choices may depend on $u$ without being measurable in $u$. The function to be integrated is $A_N$, whose measurability has already been proved independently.

Finiteness of $B$ allows the individual label limits to be summed. For infinite $B$, the argument provides no uniform control justifying that sum. The finite-block construction is therefore essential.

Since $0\le A_N(u)\le |B|\|f\|_{\infty }$ and $\nu$ is a probability measure, bounded convergence gives $\int A_N\,\mathrm d\nu\to0$. On the other hand, for every finite $N$, integration commutes with the finite sum:

This contradiction joins Proposition 5.5 to Proposition 6.1.

Although (61) gives Cesàro cancellation in time, R still seeks convergence for every configuration whose time gaps tend to infinity. The average rules out the nonzero witness produced by a hypothetical least failure order. It does not weaken the target to mixing only in an averaged sense.

8. Order of limits and conditions for integration

In the zero-entropy lemma, first fix $\varepsilon$, the deterministic word, and the phases. Obtain an exponential estimate uniform in those coefficients, then take a union bound over finitely many likely words. Finally let $n\to \infty$ and then $\varepsilon \downarrow 0$. No independence of the labels after conditioning is assumed.

In the algebraic lemma, first fix a label $l$. A finite $\pm 1$ relation with unit coefficients at both ends gives a finitely generated $K$. The parameters $p,d$ are then fixed before taking the residue-class limit $m\to \infty$. Torsion requires retaining the extra application of $N$.

In the polynomial lemma, induction starts with $p$, $F$, and $P$ fixed. At each step, fix $h$ for the correlation estimate as $n\to \infty$, then fix $H$ to expand the finite sum of lags, and finally let $H\to \infty$. No unproved uniform mixing estimate is applied to tests varying with $n$.

On returning to $u$, convergence is needed only in each typical fibre. The finite-block sum $A_N$ is measurably defined and uniformly bounded; individual coefficients under possibly nonmeasurable abstract identifications need not be measurably defined in the parameter. A mixing rate uniform in $u$ is unnecessary.